Problem 83 - College Algebra and Calculus: An Applied Approach

Question

Total Revenue The demand equation for a product is \(p=36-0.0003 x\), where \(p\) is the price per unit and \(x\) is the number of units sold. The total revenue \(R\) for selling \(x\) units is given by \(R=x p=x(36-0.0003 x) .\) How many units must be sold to produce a revenue of \(\$ 1,080,000 ?\)

Step-by-Step Solution

Answer

To generate a total revenue of $1,080,000, approximately 200,000 units must be sold.

Step 1: Set up the Equation

First, set up the equation for the total revenue \(R\) in terms of \(x\). The total revenue \(R\) is given by the equation \(R=x p=x(36-0.0003 x)\). The goal is to find the value of \(x\) that produces a revenue of $1,080,000. So, you substitute \(R\) with $1,080,000 in the equation: \[ x(36-0.0003 x) = 1080000 \]

Step 2: Simplify the Equation

Next, simplify this equation by distributing \(x\) across the terms in the parentheses to create a quadratic equation: \(36x - 0.0003 x^2 = 1080000\). To make it easier to solve, you rearrange the quadratic equation, placing all terms on the left-hand side, so that it can be solved: \[0.0003x^2-36x + 1080000 = 0\]. And for the ease of calculations, modify this equation further by multiplying it by \(10^4\), which gives: \[3x^2 - 3600x + 108000000 = 0\]

Step 3: Solve for x

Finally, solve for \(x\) using the quadratic formula \(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a = 3\), \(b = -3600\) and \(c = 108000000\). Substituting these values into the formula gives: \[x=\frac{3600\pm\sqrt{(-3600)^2-4*3*108000000}}{2*3}\] Solving this expression will give two possible values of \(x\). The Revenue can't be negative therefore pick the positive value.

Key Concepts

Demand Equation

To understand total revenue problems, we first need to grasp the demand equation. In this exercise, the demand equation is given by \( p = 36 - 0.0003x \). This equation helps you link the price per unit, \( p \), with the number of units sold, \( x \). Essentially, it tells us that the price decreases slightly as more units are sold. This reflects a typical market scenario where demand decreases at higher quantities, driving prices down.
It's crucial to notice that the slope of \(-0.0003\) is negative, indicating the price goes lower as

• the number of units sold increases;
• the demand drops, hence price needs to decrease to sell more units.

By setting a balance between price and units sold, businesses can optimize their revenue effectively.

Quadratic Formula

When solving for the number of units that will produce a specific revenue, in this case $1,080,000, we often encounter quadratic equations. These are equations of the form \( ax^2 + bx + c = 0 \). In our scenario, the equation becomes \( 3x^2 - 3600x + 108000000 = 0 \) after substitution and simplification.
To solve quadratics, we use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula is particularly useful when equations are not easily factorable. Here,

• \( a = 3 \)
• \( b = -3600 \)
• \( c = 108000000 \)

Calculating this will yield two potential solutions for \( x \), but only positive values are meaningful in the context of units sold.

Revenue Calculation

Total revenue is a crucial metric for businesses to gauge their financial health. In bookkeeping terms, it's simply the total amount of money generated from sales, calculated as the number of units sold \( x \) multiplied by the price per unit \( p \).
For this problem, the revenue equation is expressed as \( R = xp = x(36-0.0003x) \). This equation reflects that selling more units might lower the price per unit due to the demand relation, thus affecting the total revenue. The exercise demonstrates the application of the revenue equation to find the exact number of units needed to achieve a specific total revenue of $1,080,000.
This calculation is vital for setting sales goals and understanding the price-demand relationship, ultimately optimizing revenue. Businesses use this understanding to adjust production and pricing strategies.

Units Sold

The concluding part of the exercise involves determining the exact number of units that need to be sold to reach a particular revenue threshold. Once the quadratic equation \(3x^2 - 3600x + 108000000 = 0\) is solved using the quadratic formula, you come up with potential solutions.
Choosing the correct solution, in this case, the positive value, will provide the number of units that will yield $1,080,000 in revenue. This is important as negative units sold don't make practical sense in real-world applications.
This exercise teaches that optimizing sales involves balancing the total number of units and fluctuating prices dictated by market demand. Understanding how these elements interact is crucial for businesses to strategically achieve financial goals.